Fluid properties: density, viscosity, compressibility, surface tension

Why it matters

Every aerodynamic and hydraulic calculation starts from four properties of the fluid: how heavy it is (density), how strongly it resists shearing (viscosity), how much it squeezes under pressure (compressibility) and how its free surface pulls together (surface tension). Lift and drag scale with air density, skin friction and the Reynolds number depend on viscosity, Mach-number effects come from compressibility, and fuel atomisation and capillary rise in instrument lines come from surface tension.

Key ideas

Continuum hypothesis. We treat a fluid as continuous, so properties like ρ and p have a value at every point. This holds when the mean free path of the molecules is much smaller than the body size (Knudsen number Kn = λ/L ≪ 1). It fails in rarefied flow at very high altitude, which is why re-entry aerodynamics at 100 km needs different methods.

Density and specific weight. Density ρ is mass per unit volume. Specific weight γ = ρg is weight per unit volume, and relative density (specific gravity) is ρ divided by the density of water at 4 °C (1000 kg/m³). For a gas, ρ follows from the ideal-gas law p = ρRT with R = 287 J/(kg·K) for air. At ISA sea level (101 325 Pa, 288.15 K), ρ ≈ 1.225 kg/m³.

Viscosity. A fluid is a substance that deforms continuously under any shear stress, however small. In a Newtonian fluid the shear stress is proportional to the rate of shear strain, τ = μ du/dy. The constant μ is the dynamic viscosity (Pa·s). The ratio ν = μ/ρ is the kinematic viscosity (m²/s), which is what appears in the Reynolds number Re = VL/ν. Temperature behaves oppositely in liquids and gases:

• Liquids: viscosity comes from intermolecular cohesion, which weakens as temperature rises, so μ falls with temperature.
• Gases: viscosity comes from molecular momentum exchange, which rises with temperature, so μ rises (Sutherland's law describes this for air). Non-Newtonian fluids (blood, paints, slurries) have an apparent viscosity that depends on shear rate. The no-slip condition says fluid in contact with a wall moves with the wall; this is the origin of velocity gradients and hence boundary layers.

Compressibility and bulk modulus. Compressibility β is the fractional decrease in volume per unit rise in pressure; its reciprocal is the bulk modulus K. For water K ≈ 2.2 GPa, so it is practically incompressible. For a gas, K depends on the process: K = p for isothermal and K = γp for isentropic compression. The speed of sound is a = √(K/ρ); for an ideal gas undergoing isentropic sound waves this becomes a = √(γRT), about 340 m/s at sea level. A flow can be treated as incompressible when the Mach number M = V/a is below about 0.3, because density then changes by less than about 5%.

Surface tension. At a liquid–gas interface the molecules are pulled inward unequally, so the surface behaves like a stretched membrane with tension σ (N/m), equivalently energy per unit area (J/m²). Consequences:

• Pressure inside a curved surface exceeds outside pressure (Young–Laplace).
• Capillary rise or fall in thin tubes, depending on the contact angle θ (θ < 90° wets the wall and rises, like water on clean glass; θ > 90°, like mercury on glass, depresses). σ decreases with temperature and with surfactants. For water–air at 20 °C, σ ≈ 0.073 N/m.

Vapour pressure (closely related) sets the onset of cavitation: when local pressure drops below the vapour pressure, a liquid boils locally, which damages pumps and propellers.

Formulas

ρ = m / V and γ = ρ·g and SG = ρ / ρ_water

• ρ: density (kg/m³); m: mass (kg); V: volume (m³); γ: specific weight (N/m³); g = 9.81 m/s²; SG: dimensionless.

p = ρ·R·T

• p: absolute pressure (Pa); R: specific gas constant (287 J/(kg·K) for air); T: absolute temperature (K). Ideal gas only.

τ = μ·(du/dy) and ν = μ / ρ

• τ: shear stress (Pa); μ: dynamic viscosity (Pa·s); du/dy: velocity gradient (s⁻¹); ν: kinematic viscosity (m²/s). Newtonian fluids. For a thin film of thickness h between a fixed and a moving plate (speed U), linear profile: τ = μ·U / h.
• Units: 1 Pa·s = 10 poise = 1000 cP; 1 stokes = 10⁻⁴ m²/s.

β = −(1/V)·(dV/dp) and K = 1/β = −V·(dp/dV) = ρ·(dp/dρ)

• β: compressibility (Pa⁻¹); K: bulk modulus (Pa). For finite changes use K ≈ −Δp / (ΔV/V).

a = √(K/ρ) and for an ideal gas a = √(γ·R·T)

• a: speed of sound (m/s); γ: ratio of specific heats (1.4 for air).

Δp = 2σ / r (droplet or liquid jet sphere), Δp = 4σ / r (soap bubble, two surfaces), Δp = σ / r (cylindrical jet)

• Δp: pressure inside minus outside (Pa); σ: surface tension (N/m); r: radius (m).

h = 4σ·cosθ / (ρ·g·d)

• h: capillary rise (m, negative means depression); θ: contact angle; d: tube diameter (m).

Worked examples

Example 1 (standard): oil film under a sliding plate. Given: plate area A = 0.25 m², speed U = 2 m/s, oil film h = 0.5 mm, μ = 0.29 Pa·s (given). Find the force and power needed.

1. Linear profile: τ = μ·U / h = 0.29 × 2 / 0.0005 = 1160 Pa.
2. Force: F = τ·A = 1160 × 0.25 = 290 N.
3. Power: P = F·U = 290 × 2 = 580 W. Answer: F = 290 N, P = 580 W.

Example 2 (GATE level): capillary rise and compressibility. (a) A clean glass tube of diameter 1 mm stands in water (σ = 0.073 N/m, θ = 0°, ρ = 1000 kg/m³). Find the rise.

1. h = 4σ·cosθ / (ρ·g·d) = 4 × 0.073 × 1 / (1000 × 9.81 × 0.001).
2. h = 0.292 / 9.81 = 0.0298 m. Answer: h ≈ 29.8 mm.

(b) Water (K = 2.2 GPa, ρ = 1000 kg/m³) is pressurised by Δp = 10 MPa. Find the percentage volume reduction and the speed of sound.

1. ΔV/V = −Δp / K = −10×10⁶ / 2.2×10⁹ = −0.004545.
2. Volume falls by 0.45 %.
3. a = √(K/ρ) = √(2.2×10⁹ / 1000) = 1483 m/s.

(c) Check air at ISA sea level: ρ = p/(R·T) = 101 325 / (287 × 288.15) = 1.225 kg/m³, and a = √(1.4 × 287 × 288.15) = 340.3 m/s. With μ = 1.789×10⁻⁵ Pa·s, ν = μ/ρ = 1.46×10⁻⁵ m²/s.

Common mistakes

• Mixing up μ and ν: Re = ρVL/μ = VL/ν. Using μ where ν is needed is off by a factor of ρ (about 1000 for water).
• Saying "viscosity decreases with temperature" for all fluids. It is true for liquids and false for gases.
• Using gauge pressure in p = ρRT. The ideal-gas law needs absolute pressure and temperature in kelvin.
• Using 2σ/r for a soap bubble. A bubble has two surfaces, so Δp = 4σ/r.
• Forgetting the cosθ in capillary rise, or using radius where the formula has diameter (factor of 2 error).
• Calling a gas flow compressible just because the fluid is a gas. Below M ≈ 0.3 air flow is treated as incompressible.
• Unit slips: 1 cP = 10⁻³ Pa·s, 1 mm = 10⁻³ m, 1 GPa = 10⁹ Pa.

For GATE AE

Expect short numericals on the Newtonian shear law (plate on an oil film, rotating cylinder viscometers), conversions between dynamic and kinematic viscosity, ideal-gas density and speed of sound at a given temperature, bulk modulus and percentage volume change, and capillary rise or droplet pressure. Conceptual MCQs test the temperature dependence of viscosity in liquids versus gases, the definition of a fluid, the no-slip condition and the incompressibility limit M < 0.3. Practise quick unit handling, because these questions are usually lost on factors of 10³.

Quick check

1. Air at 288 K has μ = 1.79×10⁻⁵ Pa·s and ρ = 1.225 kg/m³. What is ν?
2. Does the viscosity of air increase or decrease when it is heated?
3. What is the excess pressure inside a soap bubble of radius 10 mm if σ = 0.03 N/m?
4. Below roughly what Mach number can air flow be treated as incompressible?
5. Convert 0.89 cP to Pa·s.

Answers: 1. 1.46×10⁻⁵ m²/s 2. Increases 3. Δp = 4σ/r = 12 Pa 4. About 0.3 5. 8.9×10⁻⁴ Pa·s